LED's more efficient than CFL's?

[Roland Paterson-Jones]

I posed this question some time back

http://groups.google.com/groups?hl=...5&[email protected]#link1

However, I have finally taken the trouble to work the numbers myself for
Nichia model NSPL510S, which is nominally rated at 2500 mcd (typical) and
2*phi-half of 50 degrees.

The conservative calculation (see below) gives 85 lumens/watt, which
surprised me hugely. Typical incandescent is apparently 17 lumens/W by
comparison, and CFL's apparently about 4 times that, i.e. about 70 lumens/W.

Can someone please check my logic and workings.

Data is from http://www.nichia.co.jp/specification/led_lamp/NSPL510S.pdf.
You will need to register to download the file, but it is free.

I used the directivity graph to plot the minimum (relative) intensity
according to half-angle for various discreet angles
(5,10,15,20,25,30,40,45,60,75). I ignored angles beyond 75 degrees, since
the intensity dropped below 2%.

Then, I used the formula, 2*PI()*(1-COS(PHI/180*PI())) in Excel to calculate
the steradians within a beam of half-angle PHI. Note that 'PHI/180*PI()'
simply converts from degrees to radians.

The steradians covered by each band (e.g. between 5 and 10 degrees) is then
simply the difference between the 5 and 10 degree total steradians.

Then I used the formula lumens = candelas * steradians to calculate the
total lumens output in each band. This is a conservative figure, since I use
the minimum intensity for each band.

Then the total lumens output of the LED is the sum of the lumens over all
bands. Here are the precise workings in CSV (comma-separated) - import it
into your favourite spreadsheet for a better view:

,Model,NSPL510S,1/2-Angle,Intensity,,Steradians,diff,Lumens,(cum)
,2phi(1/2),50,5,95%,,0.02,0.02,0.20,0.20
,Luminosity (mcd),2500,10,95%,,0.10,0.07,0.61,0.82
,Current (mA),20,15,98%,,0.21,0.12,1.05,1.86
,Voltage (V),3.6,20,80%,,0.38,0.16,1.19,3.05
,,,25,50%,,0.59,0.21,0.94,3.99
,Power (W),0.072,30,25%,,0.84,0.25,0.57,4.56
,,,40,10%,,1.47,0.63,0.57,5.13
,Lumens/W,85.22159065,45,8%,,1.84,0.37,0.27,5.39
,,,60,4%,,3.14,1.30,0.47,5.86
,,,75,2%,,4.66,1.52,0.27,6.14
,,,,,,,,,
,,,,,,,Total,6.14,

Note the total lumens is 6.14, and for a typical power rating of (20mA *
3.6V = ) 0.072W, this represents a little over 85 lumens/W.

The last column is the cumulative lumens output from the centre of the beam
out. Note that a little less than 2/3 of the light is generated within
phi-half (3.99 lumens in 25 degrees).

Comments?
Roland
--
Roland and Lisa Paterson-Jones
Forest Lodge, Stirrup Lane, Hout Bay
http://www.rolandpj.com/forest-lodge
mobile: +27 72 386 8045
e-mail: [email protected]

 

[Nick Birrer]

I posed this question some time back

http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&
frame=right&th=371dca80c7976665&seekm=b8kf9e%24ou7%241%40ctb
-nnrp2.saix.net#link1

However, I have finally taken the trouble to work the
numbers myself for Nichia model NSPL510S, which is
nominally rated at 2500 mcd (typical) and 2*phi-half of 50
degrees.

The conservative calculation (see below) gives 85
lumens/watt, which surprised me hugely. Typical
incandescent is apparently 17 lumens/W by comparison, and
CFL's apparently about 4 times that, i.e. about 70
lumens/W.

Can someone please check my logic and workings.

Data is from
http://www.nichia.co.jp/specification/led_lamp/NSPL510S.pdf.
You will need to register to download the file, but it is
free.

I used the directivity graph to plot the minimum (relative)
intensity according to half-angle for various discreet
angles (5,10,15,20,25,30,40,45,60,75). I ignored angles
beyond 75 degrees, since the intensity dropped below 2%.

Then, I used the formula, 2*PI()*(1-COS(PHI/180*PI())) in
Excel to calculate the steradians within a beam of
half-angle PHI. Note that 'PHI/180*PI()' simply converts
from degrees to radians.

The steradians covered by each band (e.g. between 5 and 10
degrees) is then simply the difference between the 5 and 10
degree total steradians.

Then I used the formula lumens = candelas * steradians to
calculate the total lumens output in each band. This is a
conservative figure, since I use the minimum intensity for
each band.

Then the total lumens output of the LED is the sum of the
lumens over all bands. Here are the precise workings in CSV
(comma-separated) - import it into your favourite
spreadsheet for a better view:

,Model,NSPL510S,1/2-Angle,Intensity,,Steradians,diff,Lumens,
(cum) ,2phi(1/2),50,5,95%,,0.02,0.02,0.20,0.20
,Luminosity (mcd),2500,10,95%,,0.10,0.07,0.61,0.82
,Current (mA),20,15,98%,,0.21,0.12,1.05,1.86
,Voltage (V),3.6,20,80%,,0.38,0.16,1.19,3.05
,,,25,50%,,0.59,0.21,0.94,3.99
,Power (W),0.072,30,25%,,0.84,0.25,0.57,4.56
,,,40,10%,,1.47,0.63,0.57,5.13
,Lumens/W,85.22159065,45,8%,,1.84,0.37,0.27,5.39
,,,60,4%,,3.14,1.30,0.47,5.86
,,,75,2%,,4.66,1.52,0.27,6.14
,,,,,,,,,
,,,,,,,Total,6.14,

Note the total lumens is 6.14, and for a typical power
rating of (20mA * 3.6V = ) 0.072W, this represents a little
over 85 lumens/W.

The last column is the cumulative lumens output from the
centre of the beam out. Note that a little less than 2/3 of
the light is generated within phi-half (3.99 lumens in 25
degrees).

Comments?
Roland

I imported your data into Excel and verified the sr/band, but
the values I got for lumens/band were considerably less.
Here's your csv with a few columns added on:

Model,NSPL510S,1/2-Angle,Intensity,,Steradians,diff,Lumens,
(cum),,sr,lumens,cumul
,,0,,,,,,,,,,
2phi(1/2),50,5,0.95,,0.02,0.02,0.2,0.2,,0.024,0.057,0.057
Luminosity
(mcd),2500,10,0.95,,0.1,0.07,0.61,0.82,,0.072,0.170,0.227
Current
(mA),20,15,0.98,,0.21,0.12,1.05,1.86,,0.119,0.291,0.517
Voltage (V),3.6,20,0.8,,0.38,0.16,1.19,3.05,,0.165,0.330,0.847
,,25,0.5,,0.59,0.21,0.94,3.99,,0.210,0.262,1.109
Power
(W),0.072,30,0.25,,0.84,0.25,0.57,4.56,,0.253,0.158,1.267
,,40,0.1,,1.47,0.63,0.57,5.13,,0.628,0.157,1.424
Lumens/W,85.22159065,45,0.08,,1.84,0.37,0.27,5.39,,0.370,0.074
,1.499
,,60,0.04,,3.14,1.3,0.47,5.86,,1.301,0.130,1.629
,,75,0.02,,4.66,1.52,0.27,6.14,,1.515,0.076,1.704

,,,,,,Total,6.14,,,,total,23.673

What are we doing different to calculate lumens? I'm using

2.5*(sr/band)*(pct intensity).

I get a total of 1.7 lumens and 23.7 lumens/watt

Nick

 

[Nick Birrer]

....

I took 9000 mcd for the WW503YJ-C as typical, and used
their 'radiation diagram' etc. from
http://www.led-center.com/files/WW503YJ-C.pdf.

That explains the 3.6x discrepancy!

This gave me 7.25 lumens total, and 69 lumens/W. However
2/3 of the lumens came from outside phi-half, which was a
little suspicious.

I looked at your %intensity estimates and I get about the same
looking at Wilycon's chart. Although the intensity is a lot
lower at high angles, the solid angle subtended by the
corresponding 'band' is a lot larger which would explain the
suspicious lumens. I'm curious about the 30ma you used for the
current - 20ma would yield an even more inflated lumen value.

A couple of notes on the lumen estimate-

In order for the estimate to be a valid lower bound
(conservative), the radiation pattern has to be axially
symmetric. I've looked at some LEDs with a magnifier and noted
that the emission region is square which suggests lack of
axial symmetry. I'm not an LED expert, but I assume this also
applies to the LEDs in question.

The other necessity for this estimate to be a valid lower
bound is that the intensity monotonically decrease with
increasing off-axis angle. This appears to be the case from
the intensity vs. angle slice shown in the .pdf, but again,
not being an expert here I'll leave it for others to verify
whether this is true or not.

Incidently, the Nichia values don't sound that impressive
compared to the figures for other ultra-white LED's.

I didn't check the Nichia specs - didn't wanna play the silly
register game. However, the numbers seem more in line with
what I would expect given what I've read about LED luminous
efficiency.

Roland

Nick

P.S. It looks like my news software wrapped some of the lines
in the .csv I previously posted. Apologies for the hassle if
anyone tried to import this.

 

[Nick Birrer]

Doh! Let's try this bit again-

you used for the current - 20ma would yield an even more
inflated lumen value.

^^^^^
How about lumen/watt instead...

 

[Victor Roberts]

Incidently, the Nichia values don't sound that impressive compared to the
figures for other ultra-white LED's.

23.7 lm/W is a very respectable number for white LEDs.